Question

\(\quad\) Suppose \(V=\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right\\} .\) If \(\mathbf{u}=a_{1} \mathbf{v}_{1}+a_{2} \mathbf{v}_{2}+\cdots+a_{n} \mathbf{v}_{n}\) where the \(a_{i}\) are in \(\mathbb{R}\) and \(a_{1} \neq 0,\) show that \(V=\operatorname{span}\left\\{\mathbf{u}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right\\} .\)

Short answer

Replace \(\mathbf{v}_1\) with \(\mathbf{u}\) to show both spans are equivalent.

Step-by-step solution

Understand the Problem

We want to show that a vector space spanned by vectors \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\) can also be expressed using a slightly different set of vectors: \(\{\mathbf{u}, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\), given that \(\mathbf{u}\) is a linear combination of these vectors and \(a_1 eq 0\).

Express \(\mathbf{v}_1\) in terms of \(\mathbf{u}\)

Since \(\mathbf{u} = a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n\), we can solve for \(\mathbf{v}_1\). Rearrange the equation as \(a_1 \mathbf{v}_1 = \mathbf{u} - a_2 \mathbf{v}_2 - \cdots - a_n \mathbf{v}_n\). Then, express \(\mathbf{v}_1\) as \(\mathbf{v}_1 = \frac{1}{a_1} \left( \mathbf{u} - a_2 \mathbf{v}_2 - \cdots - a_n \mathbf{v}_n \right)\).

Substitute \(\mathbf{v}_1\) and Show Equivalence

Substitute \(\mathbf{v}_1\) in the original span \(V = \operatorname{span}\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\). Replace \(\mathbf{v}_1\) with the expression found in Step 2. This gives \(V = \operatorname{span}\{\frac{1}{a_1}(\mathbf{u} - a_2 \mathbf{v}_2 - \cdots - a_n \mathbf{v}_n), \mathbf{v}_2, \ldots, \mathbf{v}_n\}\). Since scaling and adding vectors don't change span, it can be simplified to \(V = \operatorname{span}\{\mathbf{u}, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\).

Concluding the Equivalence

Since we can express \(\mathbf{v}_1\) in terms of \(\mathbf{u}, \mathbf{v}_2, \ldots, \mathbf{v}_n\), and all vectors from the original span are preserved, the span of the new set {\mathbf{u}, \mathbf{v}_2, \ldots, \mathbf{v}_n\} matches the span of the original set \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\}. Thus, \(V = \operatorname{span}\{\mathbf{u}, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\).

Key concepts

Vector Spaces

A vector space is a fundamental concept in linear algebra where a collection of vectors can be added together and multiplied by scalars to produce another vector within the same space.
Imagine a broad, flat plane that stretches infinitely in all directions. Any vector in this plane can be described as an addition or scalar multiplication of a few key vectors. These key vectors form a basis.
Vector spaces have specific properties: they must be closed under addition and scalar multiplication, meaning that performing these operations with vectors in the space will not produce a vector outside the space. This provides structure and predictability that mathematicians and scientists rely on extensively, especially in areas like physics and engineering where vector spaces model real-world situations.

Spanning Sets

A spanning set of a vector space is a collection of vectors such that any vector in the space can be represented as a linear combination of this set.
Think of a spanning set as the building blocks of a space: you should be able to mix them and multiply them (by scalars) to form any vector in that space. If you change one of these building blocks, such as in the exercise where \(\mathbf{v}_1\) is replaced by \(\mathbf{u}\), the spanning capacity remains intact as long as you correctly adjust the combination of vectors.
In linear algebra, adjusting the vector arrangement while maintaining their spanning capability is crucial. This is because the essence of the entire space is preserved, allowing the span of vectors like \(\mathbf{u}, \mathbf{v}_2,\ldots, \mathbf{v}_n\) to remain consistent with the original set \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\).

Linear Combinations

Linear combinations are foundational in understanding vector spaces and spans. A linear combination involves taking vectors and scaling them with real numbers (scalars), then adding the results.
In our exercise, \(\mathbf{u}\) is regarded as a linear combination of the existing vectors \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\). It is expressed as \(\mathbf{u} = a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n\). This formulation means \(\mathbf{u}\) relies on these vectors scaled by particular coefficients.
Linear combinations are useful because they let us express a potentially complex vector in terms of a simpler or more familiar set. This is extensively used in problem-solving to transform the span or redefine a vector space without losing any functional integrity. As demonstrated, replacing \(\mathbf{v}_1\) with \(\mathbf{u}\), while adjusting the other vectors, keeps the overall span unchanged.